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Approximating the $\ce{Fe}$ complex as a square base pyramid ($D_\mathrm{4h}$ symmetry) with 5 $\ce{N}$ ligands, My electron count of the system gives me 16 electrons – 10 from the $2\times 5$ $\ce{N}$ electron pairs and 6 from the iron(II). Filling the molecular orbitals of the valence shells, I ended up with $\ce{4 e-}$ in the d non-bonding orbitals – for there to be 6 in the d orbitals, I am under the impression that there must be 18 electrons, as suggested by the 18 electron rule (which I am aware is a loose guideline).

Looking at depictions of the orbital diagram however, such as this one:

Depiction ofthe electronic structure of the iron(II)-heme complex

and this one:

Change in the electronic structure upon coordination of oxygen to iron(II)-heme

I've seen that there are consistently 6 electrons in these orbitals — which would be consistent with iron(0) but not iron(II) as I understand it. Can anyone help me make sense of this situation?

If I'm being a little unclear about my approximation, just look at the $\ce{Fe}$ molecular orbitals on the lefthand side of the images and subtract 2 electrons

Jan
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jovialplutonium
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  • Related: http://chemistry.stackexchange.com/questions/49940/why-how-is-blood-red-colours-of-hemoglobin/49942#49942 – Nilay Ghosh Dec 11 '16 at 08:58

1 Answers1

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I've seen that there are consistently 6 electrons in these orbitals — which would be consistent with iron(0) but not iron(II) as I understand it.

Iron is in the 8th group (or group VIIIb in older terminology). Thus, in its ground state it has eight electrons; two of which are in the 4s orbital. However, practically always once you oxidise a transition metal and enter coordination chemistry, the orbital energies shift relatively to each other and it becomes more stable to have all the energies in the d-orbitals. Thus, the typical configuration for iron(II) is not $[\ce{Ar}]\,\mathrm{3d^4\,4s^2}$ but $[\ce{Ar}]\,\mathrm{3d^6\,4s^0}$.

I am under the impression that there must be 18 electrons, as suggested by the 18 electron rule (which I am aware is a loose guideline).

I think, you’re not aware of the looseness of the 18-electron-rule guideline. Most of the time it does nothing except for stating a maximum that can occur. However, there are still many complexes that violate it; e.g. anything octahedral with zinc(II) (22 electrons), anything with copper(II) (21), octahedral nickel(II) (20) etc. It is much more valuable when going to neutral carbonyl complexes; e.g. iron typically forms pentacarbonyliron $\ce{[Fe(CO)5]}$ due to the 18 electron rule. But in the context of standard cationic complexes. you can safely ignore it.

Jan
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  • Where are the 2 electrons coming from to fill the d orbitals since 2 have been lost in the transition form iron(0) to iron(II)? Also, I don't mean that there must be 18 electrons in the sense that this is needed from a stability standpoint but rather that for there to be 6 electrons in the d orbitals is indicative of 18 electrons in the overall system.

    I guess my question is, considering the orbital diagram for D4h, it seems as though with 16 electrons we should only observe 4 electrons in these d orbitals, but these diagrams depict 6.

     – jovialplutonium Dec 11 '16 at 16:06
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    Regarding the first part of your answer, I wrote a little bit on it before: In crystal field theory, why do all the metal electrons enter the d orbitals and not the s orbital? – orthocresol Dec 11 '16 at 16:08
  • @jovialplutonium Huh? Iron(0) has eight electrons in the ground state ($[\ce{Ar}],\mathrm{3d^6,4s^2}$. Two are lost which we can assume to be the s-electrons leaving the d-electrons. If the coordination sphere is pentacoordinate, there will only be 16 electrons in total. – Jan Dec 11 '16 at 16:12
  • @orthocresol--Your response pretty much sums up my confusion. We have (by my perhaps erroneous count) 16 electrons in the system--looking at your MO diagram from your link (and keeping in mind that it is describing an octahedron, not a square-base pyramid [D4h]) and simply filling the orbitals up to 16 electrons, we should have only 4 electrons in the t2g non-bonding orbital, right? But when looking at diagrams of heme, I've seen that there are 6 electrons in these orbitals--which is indicative of 18 electrons, not 16. – jovialplutonium Dec 11 '16 at 16:14
  • @Jan I misread your answer a bit, but my question still stands. If you look at the MO diagram from the link provided by orthocresol, and fill it in with 16 electrons, it seems as though we should only observe 4 electrons in the non-bonding t2g orbitals, not 6. The orbital diagram is for an octahedron but I believe that the change in geometry wouldn't necessarily change this fact. – jovialplutonium Dec 11 '16 at 16:18
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    @jovialplutonium You need to remember to remove one ligand-centred orbital (i.e. one of the lower ones) since you only have a pentacoordinated metal centre. Thus, for pentacoordinate 16-electron systems, the central metal has six electrons in its $\mathrm{t_{2g}}$ orbitals. – Jan Dec 11 '16 at 16:23
  • @Jan That makes sense--I'll have to work through that, then.

    Thanks!

     – jovialplutonium Dec 11 '16 at 16:24
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    @jovialplutonium Yes, exactly that. If you have five ligands, you have five lower-energy ligand-based orbitals, for 10 electrons. The remaining six belong to iron. – orthocresol Dec 11 '16 at 16:26
  • Can either of you recommend any good resources about these kinds of diagrams for a variety of different point groups? – jovialplutonium Dec 11 '16 at 16:33